fusor volumetric efficiency

[guest]

This is my first attempt at the new field of fusor efficiency but here goes.

I base this on the Lawson's Criteria.
Simply take the NT of your experiment and divide by the the Lawson's Criteria then multiply by 100 to get the percent efficiency. To get a going fusion reactor system will mean results in excess of 100%.
The containment time is simply the time you have your poissor lit up.
I order to get the N for the fusor....
I picture the fusor as a sperical collector of ions at a focal point. But in order to derive an equation one must
know the neutral gas density (deuterium gas) under vacuum! A little rough to say the least. So I settled on the observed diameter of the poissor as my starting point. So from the diameter we can calculate the volume ....so far so good. But N is in units of s/m^3
,so we have aboult half of what we need.
The other missing piece is the number ions in the poissor. How the heck can we know this? By figuring out how many ions are made at what amperage is used in the fusor. For a singlely charged deuterium ion ,one electron must by removed. The charge on an electron is measured in coulombs and is fixed at 1.60 times ten to the minus nineenth. One coulomb is as defined as one Amphere second. So to get the number of ions simply divide the charge of one electron into one coulomb to get 6.25 times 10 to the eighteen ions.
So one Amphere Second equals 6.25X10^18 ions.
But we need the number of particle so we must multiply by the containment time (Tc).
Finally to get N we divide 6.25x 10^18 by the volume so we get....
......... 6.25 x 10^18a
N= ..__________________
.... 4/3 Pi D^3/8


N = 3.75 x 10^19a
.. ____________
.. D^3 Pi

so the new efficiency eq becomes.

( NT fusor / 3.0 x 10^20 ) x 100 = % fusor vol eff

(3.98x10 ^-2 A Tc^2/ D^3)x 100 = % fusor vol eff

A in amperes
Tc in seconds
D in meters

Enjoy!
Fusion is fun!

Larry Leins
Fusion Tech